RMS value of a fully rectified clamped sinusoid.

[itchy8me]

so i have a question about calculating the RMS value of a fully rectified clamped sinusoid.

Assumptions:
The top of the waveform = U
It is clamped at 0.5 U

I can calculate the RMS value by adding the 3 components of the wave, ei. @ 0.5U ω = π/6 & 5π/6 which forms a block and two side compnents. I then calculate integral of the squares of the two side components of the sinusoid and the block wave spanned from π/6 to 5π/6. and i find the RMS value which is ≈ 0.44*U.

Now i tried to calculate this a different way, but i could not get it to work. I thought it was logically equivalent to adding the components, but it is not working. Maybe someone could shed some light on my fallacies.
What i am trying to do is to calculate the RMS value of the complete rectified sinusoid and then substract the RMS value of the piece of sinusoid that has been clamped off. I found the piece that has been clamped off to be:

√(1/π * ∫(from π/6 to 5π/6) (-0.5*U + U*sin(ωt))^2 dωt) {-0.5*U drops the sinusoid so that i can find the RMS value of only the clamped of part. At least this is what i assume...}

and so i subtract that from the RMS value of the fully rectified sinusoid which is U/√2

This however provides me with a wrong answer which was somewhere around 0.27*U . What am i wrongfully assuming or calculating?

 

[Stephen Tashi]

Now i tried to calculate this a different way, but i could not get it to work. I thought it was logically equivalent to adding the components, but it is not working. Maybe someone could shed some light on my fallacies.

I haven't read all the details of your work, but In general, $\int {(f(x) + g(x))^2 dx}$ is not equal to $\int {f^2(x) dx} + \int {g^2(x) dx}$. So you can't compute the RMS by expressing a wave as a sum of arbitrary components. You must choose the components to be orthogonal functions .